JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 11)
The resistance of the meter bridge AB in given figure is 4 $$\Omega $$. With a cell of emf $$\varepsilon $$ = 0.5 V and rheostat
resistance Rh = 2 $$\Omega $$ the null point is obtained at some point J. When the cell is replaced by another one of emf $$\varepsilon $$ = $$\varepsilon $$2 the same null point J is found for Rh = 6 $$\Omega $$. The emf $$\varepsilon $$2 is, :
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0.3 V
0.6 V
0.5 V
0.4 V
Explanation
Potential gradient with Rh = 2$$\Omega $$
is $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$$ L $$=$$ 100 cm
Let null point be at $$\ell $$ cm
thus $$\varepsilon $$1 $$=$$ 0.5V $$=$$ $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell $$ . . .(1)
Now with Rh $$=$$ 6$$\Omega $$ new potential gradient is
$$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$$ and at null point
$$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$$ . . .(2)
dividing equation (1) by (2) we get
$${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$$ thus $${\varepsilon _2} = 0.3$$
is $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$$ L $$=$$ 100 cm
Let null point be at $$\ell $$ cm
thus $$\varepsilon $$1 $$=$$ 0.5V $$=$$ $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell $$ . . .(1)
Now with Rh $$=$$ 6$$\Omega $$ new potential gradient is
$$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$$ and at null point
$$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$$ . . .(2)
dividing equation (1) by (2) we get
$${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$$ thus $${\varepsilon _2} = 0.3$$
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